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I'm about to buy a pair of headphones and an audio interface with built in amplifier. The specs say that the amp's impedance is "<30 Ohms".

The headphone I'd like to buy is a Beyerdynamic DT 990, which has versions of different impedance.

I'm only qualified in electronics enough to know that the higher the headphone impedance, the more "amplification" (for a lack of a better word) required to achieve the same power.

However, I'm concerned that the impedance being significantly different introduces distortion to the sound. I'm not talking about saturation necessarily, but maybe a slight change in transfer characteristics, which is obviously not something I want to deal with.

Any insight on this topic is highly appreciated.

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    \$\begingroup\$ Depends on if you want a realistic or an audiophool level answer. I personally try things out, and if I like the sound I use it. \$\endgroup\$ – PlasmaHH Aug 31 '18 at 8:04
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    \$\begingroup\$ Please do a simulation with real life parasitics and real life speaker over audio frequencies and sweep the source impedance. Can you see any difference? \$\endgroup\$ – winny Aug 31 '18 at 8:06
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    \$\begingroup\$ well, I asked here because I'm curious about the science part of the question. I never cared about the "I hear sg on the midrange" kind of answers. I'm interested in the science behind this. Pure math and physics. \$\endgroup\$ – László Stahorszki Aug 31 '18 at 8:23
  • \$\begingroup\$ Please see my answer to the same question, though it is more about power than impedance. electronics.stackexchange.com/questions/242736/… \$\endgroup\$ – Sparky256 Sep 1 '18 at 2:04
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You can simply forget about impedance matching for home Audio.

Impedance matching is needed only where the wavelength of the signal comes close to the length of the cable transporting that signal. Electrical signals travel with almost the speed of light through cables, for the highest audio frequency (giving the shortest wavelength) the wavelength is about 15 km. I'm guessing that your cables aren't that long.

Impedance matching is needed to prevent signals reflecting and distorting them. This is usually only relevant for high frequency signals, not audio (exception: analog telephone lines).

In my opinion the "impedance matching" for audio amplifiers is really better understood as: "Can this amplifier drive this speaker?"

Example: some amplifiers are only suitable for 4 and 8 ohm speakers. Using it with 2 ohm speakers (or two 4 ohm speakers in parallel) can give issues.

For headphones this is almost never an issue unless the impedance of the headphone is very low (less than 10 ohms) or very high (600 ohms). And even then, if there is a "mismatch" the maximum volume might be reduced.

Usually home audio amplifiers drive the headphone output from the speaker output via series resistors to give a bit of protection against overloading the headphones as they need a lot less power than the speakers. Because of this almost any headphone can be driver from a home audio amplifier.

Mobile devices running on batteries cannot deliver so much power and voltage so overloading is less of an issue. Since the output voltage on these devices is limited I recommend using low impedance headphones, 30 or 50 ohms would be a good choice.

In either case, you do not have to worry about impedance matching, it is really a non-issue for headphones.

Sidenote:

For speakers the output impedance of the amplifier is relevant. The usual recommendation is that the amplifier needs a low output impedance. The lower the better as that will give it better "control" over the speaker. This is not impedance matching, it is actually a "best mismatch" situation as amplifier output impedance ( < 0.1 ohms) and speaker impedance (> 4 ohms) are not the same.

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  • \$\begingroup\$ Regarding the sidenote, the ratio between amp output impedance and speaker impedance is called the damping factor. en.wikipedia.org/wiki/Damping_factor \$\endgroup\$ – Dampmaskin Aug 31 '18 at 9:49
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    \$\begingroup\$ The OP seems to be saying that the amplifier's output impedance is "<30 ohms" so why do you assume it is "<0.1 ohms"? Doesn't maximum power transfer from a Thevenin equivalent circuit come into play here? The maximum power transfer is when the load impedance is equal to the source's output impedance. \$\endgroup\$ – Elliot Alderson Aug 31 '18 at 12:10
  • \$\begingroup\$ @ElliotAlderson That "< 30 ohms" is vague to me so I did not discuss it. I am 100% sure that a headphone requiring "Rout < 30 ohms" will also work just fine with an amplifier with a 100 ohms output impedance, provided it can make enough output voltage so that the signal power in the headphone is sufficient. That <0.1 ohms is a typical number I sucked out of my thumb. Most audio amps. have an Rout< 0.1 ohms. Given they have feedback that is not a challenge to design for that. \$\endgroup\$ – Bimpelrekkie Aug 31 '18 at 14:02
  • \$\begingroup\$ OK, but I think you should have included those assumptions in your answer. \$\endgroup\$ – Elliot Alderson Aug 31 '18 at 14:05
  • \$\begingroup\$ @ElliotAlderson You're right about Thevenin but that kind of power matching does not work for an Audio amp+speaker. Audio Amps have a low Rout so that they approach the behavior of a voltage source. An ideal voltage source can deliver all the power you want. An audio amp cannot. The maximum voltage and current are designed to handle 2 to 8 ohms. If you would use a value of Rout as per Thevenin then the current would be the limiting factor, the audio amp cannot deliver so much current. \$\endgroup\$ – Bimpelrekkie Aug 31 '18 at 14:06
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You can certainly forget about "impedance matching" for any reasonable quality amplifier, because the impedances are intentionally not matched!

The output impedance of a good quality audio amp should be very low - typically something like 0.01 or even 0.001 ohms. The impedance of most loudspeakers or headphones varies a lot with frequency, but is always several orders of magnitude higher than the amp's output impedance.

The consequence is that if the speaker is designed to respond to the signal voltage (not current), the voltage delivered by the amp will not depend on the impedance variations of any particular make of loudspeaker, and the audio response won't have any unwanted peaks or troughs at particular frequencies in the audio range.

Impedance matching only matters if you need to get the maximum possible power transfer from one device to another, but that is irrelevant for designing an audio amp. (But if you were trying to send a signal down a few thousand miles of undersea cable, it would be very relevant!)

The reason the amp specification mentions the acceptable output impedance range for the speakers or phones is simply to ensure that when you turn the amp volume control up to the maximum, (1) if the speaker impedance is too high, the sound level will be lower than what you might expect, and (2) if the impedance is too low, you will be attempting to draw too much current from the amp, which might cause some distortion, and (more likely) will blow a fuse somewhere in the amp to prevent it being overloaded.

Historical note: the above applies to modern amp designs - both solid state and tubes (valves). Some old (1950s or 60s) designs of tube amps were sensitive to the speaker output impedance connected to them, and had a switch on the back panel to select the actual impedance being used (usually either 8 or 16 ohms in those days, though high powered modern speakers often have lower impedance like 4 or 2 ohms). Operating such a "vintage" amp with no loudspeaker connected could in some cases damage the amp - but modern designs of tube amp don't have that problem.

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  • \$\begingroup\$ those are some great answers here, can I accept multiple answers? :D \$\endgroup\$ – László Stahorszki Sep 4 '18 at 15:24
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The "< 30 Ohms" designation on your amplifier is most likely not the output impendance of the amp, but the impendance of the load it is designed for. Good audio amplifiers have output impendances much below one Ohm.

Such labeling is common, because impendance matching is not a thing in audio equipment. However, driving speakers or headphones with impendance outside the amplifier's design area might affect sound.

Driving headphones with impendance higher than amplifier's designed driven impendance is likely fine. I would believe the only drawback is that the amplifier can not deliver its full output power since it is not able to generate high-enough output voltages, and you'll need to stay at fairly low output levels. Using the volume knob may be fiddly, but amount of power is not really a concern as headphones never need much power.

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    \$\begingroup\$ “driving speakers or headphones with impedance outside the amplifier's design area might affect sound” – driving a speaker with too-low impedance might most importantly destroy the amp, or at least shut it down by the short circuit or overheat protection. \$\endgroup\$ – leftaroundabout Aug 31 '18 at 13:40
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Most of the time the accepted answer is acceptable .Tube amps have also been discussed .Well what if the amp is class D which is becoming more common .The normal Class D amp will have a LC filter between the output transistors and the speaker .Some low end low power stuff relies on the speaker inductance and very short speaker wires.The LC filter will reduce radiation from long speaker wires .The LC filter is usually set up at a low loaded Q with the cut frequency above the audio but below the PWM frequency .Ballpark numbers could be PWM 200KHz F cut 25 KHz and Q .7 .Now the speaker impedance will change filter parameters .Raising the impedance will raise the Q with most orthodox LC filter designs .This could give a horrible high end peak and possibly muck up the negative feedback loop .You may get bad performance which could be dealt with with a simple passive network .

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A typical amplifier which is designed to (fed some particular input, and at some particular volume setting) feed 50 watts into speakers with an 8 ohm impedance would typically drive 16 ohm speakers with only 25 watts under those conditions, and would "try" to drive 4 ohm speakers with 100 watts. The 16-ohm impedance typically wouldn't damage anything, nor even sound bad, but the amp wouldn't be able to feed as much power as when driving an 8-ohm load. The 4-ohm situation might not damage anything if the amp was designed to--under different conditions--feed 100 watts into 8 ohm speakers, but if a low impedance load would cause the amp would try to output more power than its intended design maximum, distortion and/or damage would likely result.

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Output stages shape the load current 'at all frequencies' that be Impedance Matching (in the frequency domain). Output circuits vary a lot depending their 'matching approach' be that part of what frequencies it has been predicted to 'match' and what frequencies (and currents) it actually works in.

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    \$\begingroup\$ Welcome to EE.SE, Diego. Your answer has a whiff of technobabble about it. Amplifiers typically have no filtering on the output but drive the speakers directly with a low impedance drive as explained in the other answers. "Kick capacitors" (whatever they are) or any other capacitors are not added in parallel with speakers on their own although they may form part of a cross-over circuit. "Coarse currents" sounds like a mis-translation but your user profile doesn't say what your native language is. You can edit your answer to improve it. \$\endgroup\$ – Transistor Aug 31 '18 at 20:03
  • \$\begingroup\$ @Transistor. Internally most amplifiers have a 10 uH air inductor in parallel with a 10 ohm 10 W resistor at the amp output. It is NOT for impedance matching, rather it is to protect the internal negative feedback loop from the capacitive loading of long speaker cables. \$\endgroup\$ – Sparky256 Sep 1 '18 at 2:11
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    \$\begingroup\$ If you want to say something useful then please do so. However this answer seems to make little sense at all. The At this point i cannot go further on its function on frequency domain to the detail. says enough, you obviously have no clue. That's fine but then don't pretend you do. There are very seasoned engineers here and they see right through that (myself included). \$\endgroup\$ – Bimpelrekkie Sep 1 '18 at 10:16
  • \$\begingroup\$ thank you for your comments. i'm not very seasoned in social interaction on this network to the point of expecting credits for a suggested reading. in this case about audio fidelity. such a seasoned expertise field. \$\endgroup\$ – Diego Cadogan Sep 1 '18 at 20:18
  • \$\begingroup\$ @DiegoCadogan, thanks for taking the effort to edit your initial answer, the changes did improve it. \$\endgroup\$ – Sz. Apr 20 at 1:04

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